When to use the binomial distribution
Use the binomial distribution when you count successes in a fixed number of trials, each trial has only two outcomes, the trials are independent, and the chance of success p is the same every time.[1][2] Coin flips, guessing on multiple-choice questions, free throws and defective parts in a batch all fit; drawing cards without replacement from a small deck does not, because p changes after each draw.
The binomial formula, step by step
P(X = x) = C(n, x) × pˣ × (1 − p)ⁿ⁻ˣC(n, x) = n! ÷ (x! (n − x)!) counts the ways to place x successes among n trials.
Worked example: 7 heads in 10 flips
C(10, 7) = 120, and 0.5⁷ × 0.5³ = 1 ÷ 1024. So P(X = 7) = 120 ÷ 1024 = 0.1172.
At least 7 heads adds 8, 9 and 10: (120 + 45 + 10 + 1) ÷ 1024 = 176 ÷ 1024 = 0.1719.
| Heads (k) | P(X = k) | P(X ≤ k) |
|---|---|---|
| 0 | 0.001 | 0.001 |
| 1 | 0.0098 | 0.0107 |
| 2 | 0.0439 | 0.0547 |
| 3 | 0.1172 | 0.1719 |
| 4 | 0.2051 | 0.377 |
| 5 | 0.2461 | 0.623 |
| 6 | 0.2051 | 0.8281 |
| 7 | 0.1172 | 0.9453 |
| 8 | 0.0439 | 0.9893 |
| 9 | 0.0098 | 0.999 |
| 10 | 0.001 | 1 |
Exactly, at most, at least: which probability?
Match the words in the question to the inequality; this is where most binomial mistakes happen. The calculator prints all five at once.
| Question says | Means | Value |
|---|---|---|
| exactly 6 | P(X = 6) | 0.1916 |
| at most 6, no more than 6 | P(X ≤ 6) | 0.608 |
| fewer than 6, less than 6 | P(X < 6) = P(X ≤ 5) | 0.4164 |
| at least 6, 6 or more | P(X ≥ 6) = 1 − P(X ≤ 5) | 0.5836 |
| more than 6 | P(X > 6) = 1 − P(X ≤ 6) | 0.392 |
On a TI-84, “exactly” is binompdf(n, p, x) and “at most” is binomcdf(n, p, x); “at least x” is 1 − binomcdf(n, p, x − 1).
Mean, standard deviation and the normal approximation
A binomial count has mean np and standard deviation √(np(1 − p)).[1] For 20 trials at 30%, that is 6 and 2.0494. When np and n(1 − p) are both at least 10, the count is close to normal, and the calculator shows the normal approximation with a continuity correction next to the exact value.[3] The exact value is always the one to report; it is computed through the incomplete beta function, so even P(X ≥ 900) for n = 1,000 keeps its digits.
Guessing on a testGuessing all 20 questions of a four-choice test, the chance of getting at least 10 right is 0.0139: about 1 in 72. Expected score from guessing: 5 of 20.
Common binomial mistakes
- Off-by-one on “at least”. P(X ≥ 6) is 1 − P(X ≤ 5), not 1 − P(X ≤ 6).
- Using a percent as p. 30% is p = 0.3; the calculator accepts either.
- Applying it to dependent trials. Without replacement from a small population, use the hypergeometric distribution.
Related: the z-score calculator for normal areas, the confidence interval calculator for a proportion, and the statistics calculator for data you have already collected.
How we check this calculator
Tests compare the calculator with exact fractions: P(X ≤ 3) for 10 fair flips must equal 176/1024, P(X = 5) must equal 252/1024, P(X ≥ 8) must equal 56/1024; the probabilities for n = 1,000 must add up to 1 within 10⁻¹².[2]
- 14 November 2026: page and calculator published.
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- P-value calculatorFrom z, t, chi-square or F
- Confidence interval calculatorMeans and proportions, t or z
- t-test calculatorOne-sample, Welch, pooled and paired
Frequently asked questions
How do you calculate a binomial probability?
Multiply the number of arrangements C(n, x) by pˣ and by (1 − p)ⁿ⁻ˣ. For 7 heads in 10 fair flips: 120 × (1/1024) = 0.1172.
How do I find the probability of at least x successes?
Use 1 − P(X ≤ x − 1). The calculator gives P(X ≥ x) directly.
What is the mean of a binomial distribution?
np: the number of trials times the chance of success. 20 trials at 30% average 6 successes.
What is the difference between binompdf and binomcdf?
binompdf gives the probability of exactly x successes; binomcdf gives the probability of x or fewer.
What are the conditions for a binomial distribution?
A fixed number of trials, two outcomes per trial, independent trials and the same probability of success on every trial.